Question

A body falling freely from a given height $H$ hits an inclined plane in its path at a height $h$. As a result of this impact the direction of the velocity of the body becomes horizontal. For what value of $(h/H)$ the body will take maximum time to reach the ground?

• A. $\frac{1}{3}$
• B. $\frac{1}{2}$
• C. $\frac{2}{5}$
• D. $\frac{2}{3}$

Answer 1

Answer: (B)

$PQR$ is an inclined plane. The body falls under gravity vertically from $A$ to $B$ till it strikes the inclined plane. At $B$, its velocity is horizontal. From $B$ to $C$, the body follows a parabolic path. At $C$, it reaches the ground.
Let the time taken to travel from $A$ to $B=t_1$
Time taken to travel from $B$ to $C=t_2$

$\because S=ut+\frac{1}{2}g{t}^2$
$\therefore \left(H-h\right)=\left(0\times t_1\right)+\frac{1}{2}g{t}_1^2$
or $t_1=\sqrt{\frac{2\left(H-h\right)}{g}}$
Similarly, $t_2=\sqrt{\frac{2h}{g}}$
$\therefore$ Total time, $t=t_1+t_2$
$=\sqrt{\frac{2}{g}}\left[\sqrt{\left(H-h\right)+\sqrt{h}}\right]$
$\therefore \frac{dt}{dh}=\sqrt{\frac{2}{g}}\left[\frac{-1}{2\sqrt{H-h}}+\frac{1}{2\sqrt{h}}\right]$
For $t$ to be maximum, $\frac{dt}{dh}=0$
$\therefore 0=\sqrt{\frac{2}{g}}\left[\frac{-1}{2\sqrt{H-h}}+\frac{1}{2\sqrt{h}}\right]$
or $H-h=h$
$\Rightarrow H=2h$
$\Rightarrow \frac{h}{H}=\frac{1}{2}$
$\therefore \frac{h}{H}=\frac{1}{2}$.